Aberration in Gravito-Electromagnetism
471
For simplicity, consider a heavy mass, M H connected to one of our oscillating
masses by a short (taken for convenience to be length zero), ideal spring. Then the
equilibrium position of the two masses is at the same place, say X 0 . The motion
must conserve momentum, hence
M
dΔ(t)
dt
+ M H
dΔ H (t)
dt
= 0,
(17)
where Δ(t) and Δ H (t) are the time dependent parts of the motions of the
two masses. Clearly this implies that the dipole moment p(t) is actually time
independent
d
dt
p(t) =
d
dt
M(R 0 + Δ(t)) +
d
dt
M H (R 0 + Δ H (t)) = 0.
(18)
Even though
dΔ H (t)
dt
= −
M
M H
dΔ(t)
dt
λ
dΔ(t)
dt
(19)
as
M
M H
→ 0, however, we still have
M
d 2 Δ(t)
dt 2 + M H
d 2 Δ H (t)
dt 2
= 0,
(20)
i.e.,
Ma + M H a H = 0
(21)
for the acceleration. Thus our result Eq. (16) for each of our oscillating masses is
exactly cancelled by the contribution of the necessary compensating masses. We
understand that this is how it should be. However, we can now easily ascertain that
the similar analysis does not apply to quadrupole time dependence. The quadrupole
moment Q is constructed from
Q ∼ MR
2 .
(22)
Then
˙
Q ∼ 2MR(t) ˙
R(t) = 2MR ˙
Δ(t) = 2M(R 0 + Δ(t)) ˙
Δ(t).
(23)
Realizing Δ H (t) for the compensating heavy mass we would find
˙
Q H ∼ 2M H R H (t) ˙
R H (t) = 2M H R H (t) ˙
Δ H (t) = 2M H (R 0 + Δ H (t)) ˙
Δ H (t).
(24)